Problem: Find $\lim_{x\to 1}\dfrac{x-1}{\sqrt{5x-1}-2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{5}{2}$ (Choice B) B $\dfrac{4}{5}$ (Choice C) C $-\dfrac{1}{5}$ (Choice D) D The limit doesn't exist
Explanation: Substituting $x=1$ into $\dfrac{x-1}{\sqrt{5x-1}-2}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{x-1}{\sqrt{5x-1}-2} \\\\ &=\dfrac{x-1}{\sqrt{5x-1}-2}\cdot\dfrac{\sqrt{5x-1}+2}{\sqrt{5x-1}+2} \gray{\text{Rationalize the denominator}} \\\\ &=\dfrac{(x-1)(\sqrt{5x-1}+2)}{(5x-1)-2^2} \\\\ &=\dfrac{\cancel{(x-1)}(\sqrt{5x-1}+2)}{5\cancel{(x-1)}} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{\sqrt{5x-1}+2}{5} \text{, for }x\neq 1 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $1$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x-1}{\sqrt{5x-1}-2}=\dfrac{\sqrt{5x-1}+2}{5}$ for all $x$ -values in the interval $(0.5,1.5)$ except for $x=1$. Therefore, $\lim_{x\to 1}\dfrac{x-1}{\sqrt{5x-1}-2}=\lim_{x\to 1}\dfrac{\sqrt{5x-1}+2}{5}=\dfrac{4}{5}$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 1}\dfrac{x-1}{\sqrt{5x-1}-2}=\dfrac{4}{5}$.